2

我有这样的回应：

[ { "key": { "kind": "UserRecord", "id": 0, "name": "1" }, "firstName": "1", "lastName": "1", "homeLat": 0.0, "homeLon": 0.0, "workLat": 0.0, "workLon": 0.0, "currentLat": 10.0, "currentLon": 10.82, "timestamp": 1335735046606, "score": 0, "isStarred": false, "distance": 0.0 }, { "key": { "kind": "UserRecord", "id": 0, "name": "32423542324234324" }, "firstName": "Simone", "lastName": "Boscolo Berto", "homeLat": 0.0, "homeLon": 0.0, "workLat": 0.0, "workLon": 0.0, "currentLat": 55.786444, "currentLon": 12.515867, "timestamp": 1335884083696, "score": 0, "isStarred": false, "distance": 0.0 } ]

以及从每个 JSONObject 中获取我的对象的方法

private User getUserFromJson(JSONObject jsonUser)

如何在 JSONObjects 列表中进行迭代？我应该使用 JSONArrayobject 吗？

4

3 回答 3

1

使用 Gson 或 jackson 可能会很容易

Gson 示例

Gson gson = new Gson();
User[] users = gson.fromJson(<json string>, User[].class);

杰克逊的例子

ObjectMapper mapper = new ObjectMapper();  
Collection<User> users =  
    mapper.readValue(<json string>, `new TypeReference<Collection<User>>() {});

样品-

package com.test;

import com.google.gson.Gson;

public class GSonExample {
    public static void main(String[] args) {
        String json = "{\"name\":\"Duke\",\"address\":\"Menlo Park\",\"dateOfBirth\":\"Feb 1, 2000 12:00:00 AM\"}";

        Gson gson = new Gson();
        User student = gson.fromJson(json, User.class);

        System.out.println("student.getName()        = " + student.getName());
        System.out.println("student.getAddress()     = " + student.getAddress());
        System.out.println("student.getDateOfBirth() = " + student.getDateOfBirth());
    }
}

public class User {
    private String name;
    private String address;
    private String dateOfBirth;
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public String getAddress() {
        return address;
    }
    public void setAddress(String address) {
        this.address = address;
    }
    public String getDateOfBirth() {
        return dateOfBirth;
    }
    public void setDateOfBirth(String dateOfBirth) {
        this.dateOfBirth = dateOfBirth;
    }
}

类路径

<?xml version="1.0" encoding="UTF-8"?>
<classpath>
    <classpathentry kind="src" path="src"/>
    <classpathentry kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER/org.eclipse.jdt.internal.debug.ui.launcher.StandardVMType/JavaSE-1.6"/>
    <classpathentry kind="lib" path="lib/gson-2.1.jar"/>
    <classpathentry kind="output" path="bin"/>
</classpath>
于 2012-05-01T19:17:28.793 回答
0

像这样的东西：

final String incomingJSON ;
final JSONArray objArray = new JSONArray(incomingJSON);
for(int i = 0; i < objArray.length(); i++) {
    final JSONObject obj = objArray.getJSONObject(i);
    final User user = getUserFromJson(obj);
}
于 2012-05-01T18:59:49.190 回答
0

你在使用 Gson api 吗？这样，您可以从 JsonArray 对象中获取迭代器。

Iterator<JsonElement>   iterator()
          Returns an iterator to navigate the elemetns of the array.
于 2012-05-01T19:03:35.410 回答